Talk:Ackermann function
In the article said that Ackermann function is based on zeration, but isn't it true that any function based on it? As Sbiis Saibian noted, zeration is googologically elementary function, all functions eventually will be decompozed to zerations. For example, \(n^2 = n \times n = n+n+n\cdots n+n+n = n+n+n\cdots n+n+1+1+1\cdots 1+1+1\). Ikosarakt1 (talk ^ ) 20:00, April 22, 2013 (UTC) True, but the first level of the Ackermann function is explicitly the successor function. Deedlit11 (talk) 20:13, April 22, 2013 (UTC) Zeration is, if I remember correctly: \\circ a = S(S(a))\ \\circ b = b \circ a = S(a)\text{ for }a > b\ FB100Z • talk • 21:50, April 22, 2013 (UTC) The main property of zeration that it is the unary function. So, it is just \(f(n)=n+1\). The reason that \(a \uparrow^{n} a = a \uparrow^{n+1} 2\) doesn't works here, because addition has unique property that \(a+1 \neq a\), while \(a \times 1 = a^1 = a \uparrow\uparrow 1 = a \uparrow\uparrow\uparrow 1 = a \uparrow^{n} 1 = a\). Ikosarakt1 (talk ^ ) 22:03, April 22, 2013 (UTC) Maybe I'm just being pedantic, but hyper operators are binary operators. The successor function is not a hyperoperator. I guess it's open to interpretation, the math works fine either way. FB100Z • talk • 22:16, April 22, 2013 (UTC) Original definition According to , the most known variant of Ackermann function should be called Ackermann-Péter function, but originally this function was defined differently. Ikosarakt1 (talk ^ ) 21:46, April 27, 2013 (UTC) Largest perfect power of the form 2n - 3 Is 125 the largest perfect power of the form 2n - 3? -- 22:05, December 29, 2014 (UTC) :It appears that it's an open problem whether there are any perfect powers beyond 128 which differ from other perfect power by 3. Case of one power being power of 2 might simplify the (conjectured?) proof of nonexistence of further solutions, but I wouldn't be surprized if this was an open problem. LittlePeng9 (talk) 22:46, December 29, 2014 (UTC) ::A hint: No number of the form 2n - 3, where n ≥ 3, can be a square number, since all numbers of the form 2n - 3, where n ≥ 3, are congruent to 5 modulo 8, and all square numbers are congruent to 0, 1, or 4 modulo 8. -- 10:39, December 30, 2014 (UTC) :::Suppose that Eric Reyssat's triple (2 + 310 · 109 = 235, with q = 1.6299) is actually the abc triple with the highest quality. Let 2n - 3 = am, where n ≥ 3 and m ≥ 2. The hint says that m must be odd, so we have m ≥ 3. Then we have rad(3 · am · 2n) ≤ 6a, and thus 2n ≤ rad(3 · am · 2n)q ≤ (6a)q; with a3 ≤ am = 2n - 3 < 2n, it follows that a < 2n/3, and therefore 2n ≤ (6a)q < (6 · 2n/3)q = 6q · 2q/3 · n, which is equivalent to 2(1 - q/3) · n < 6q = 2ln(6)/ln(2) · q. So we have (1 - q/3) · n < ln(6)/ln(2) · q, which is equivalent to n < ln(6)/ln(2) · q/(1 - q/3) = 9.2255…; thus we have n ≤ 9, and we can check all cases explicitly; it follows that 125 would be the largest perfect power of the form 2n - 3. -- 11:30, December 30, 2014 (UTC) IMO 1981-6 Problem 6 of IMO 1981-6 used the Ackermann function. It askes to find A(4,1981). Wythagoras (talk) 09:27, February 21, 2015 (UTC) :2^^1984-3 of course Cookiefonster (talk) 13:53, February 21, 2015 (UTC) h Wouldn't it be more efficient to define A(x,0) = A(x-1,x) instead of A(x,0) = A(x-1,1)? If so, how much more efficient? How would it compare to the normal function? ArtismScrub (talk) 17:08, December 2, 2017 (UTC)